Two charges $-q$ each are separated by distance $2d$. A third charge $+q$ is kept at mid-point $O$. Find the potential energy of $+q$ as a function of small distance $x$ from $O$ due to $-q$ charges. Sketch P.E. v/s $x$ and convince yourself that the charge at $O$ is in an unstable equilibrium.

Two charges -q each are separated by distance 2d. A third charge +q is kept at mid-point O. Find the potential

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Concept understanding — Electric Potential

Electric Potential

The Idea

When you lift a book onto a shelf you do work against gravity, and the book gains gravitational potential energy that depends on where it sits. Electric charges behave the same way in an electric field. Electric potential is the electrical analogue of "height" in a gravitational field: it tells you the potential energy a unit charge would have at a given point.

Formally, the electric potential at a point is the work done in bringing a unit positive charge from infinity to that point, slowly (without giving it kinetic energy), against the electric field.

V = \frac{W}{q_0}

Here $W$ is the work done to move a small test charge $q_0$ from infinity to the point. Because both $W$ and $q_0$ are scalars, electric potential is a scalar quantity — it has magnitude but no direction. Its SI unit is the volt:

1\ \text{V} = 1\ \text{J/C}

Potential Due to a Point Charge

For a single point charge $q$ , the potential at a distance $r$ from it is

V = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r}

Notice it falls off as $1/r$ , whereas the electric field of a point charge falls off as $1/r^2$ . The potential is taken as zero at infinity, our chosen reference. A positive charge makes the potential around it positive; a negative charge makes it negative.

Potential Difference

Usually we care about the potential difference between two points $A$ and $B$ :

V_A - V_B = \frac{W_{B\to A}}{q_0}

the work per unit charge needed to move a charge from $B$ to $A$ . This is the quantity a voltmeter reads and the "voltage" that drives current in a circuit.

Relation Between Field and Potential

Field and potential are two views of the same thing. The electric field is the negative rate of change of potential with distance:

E = -\frac{dV}{dr}

The minus sign says the field points from high potential toward low potential — a positive charge, left free, rolls "downhill" in potential. Where the potential changes steeply, the field is strong.

Superposition

Because potential is a scalar, the potential due to several charges is just the algebraic sum of their individual potentials — no vectors, no angles:

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots\right)

This makes potential far easier to compute than the field, which needs vector addition. Once you have $V$ everywhere, you can get the field by differentiating.

Equipotential Surfaces

An equipotential surface is a surface on which the potential has the same value everywhere.

No work is done in moving a charge along an equipotential surface (since $\Delta V = 0$ ).
The electric field is always perpendicular to an equipotential surface. …

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The potential energy of $+q$ is $U(x)=-kq^2\!\left(\dfrac{1}{d+x}+\dfrac{1}{d-x}\right)$ , which for small $x$ behaves as $U(x)\approx-\dfrac{2kq^2}{d}-\dfrac{2kq^2}{d^3}x^2$ — a maximum at $x=0$ , so the equilibrium is unstable.

Setup. Place the two fixed charges $-q$ at $(-d,0)$ and $(+d,0)$ , and displace $+q$ a small distance $x$ along the line joining them. Its distances to the two charges are $d+x$ and $d-x$ .

1. Potential energy.

The interaction energy of $+q$ with the two $-q$ charges is

U(x)=\frac{k(+q)(-q)}{d+x}+\frac{k(+q)(-q)}{d-x}=-kq^2\left(\frac{1}{d+x}+\frac{1}{d-x}\right).

2. Expand for small $x$ .

\frac{1}{d+x}+\frac{1}{d-x}=\frac{2d}{d^2-x^2}=\frac{2}{d}\left(1+\frac{x^2}{d^2}+\cdots\right).

Hence

U(x)=-\frac{2kq^2}{d}\left(1+\frac{x^2}{d^2}+\cdots\right)=-\frac{2kq^2}{d}-\frac{2kq^2}{d^3}x^2+\cdots

The linear term vanishes by symmetry, so $x=0$ is an equilibrium ( $U'(0)=0$ ).

3. Nature of the equilibrium.

The coefficient of $x^2$ is negative, so the curve opens downward at $x=0$ . By direct differentiation,

$U''(0)=-\frac{4kq^2}{d^3}<0,$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04174 marksMCQ
$5\ m$
›Reveal solutionSolution

$W=qEd=(e)(4)(5)=20e\ \text{J}=20\ \text{eV}$ .

A proton (charge $q=e$ ) moving a distance $d$ along a uniform field $E$ has work done on it:

$W=qEd=e\times 4\ \text{NC}^{-1}\times 5\ \text{m}=20e\ \text{J}$ …
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KEAM 2026Set eng-2026-04184 marksMCQ
$+Q$
›Reveal solutionSolution

Potential is a scalar sum; each $+Q$ sits a distance $R$ from $C$ , so $V_C=2\cdot\dfrac{Q}{4\pi\epsilon_0 R}=2K$ .

In an equilateral triangle of side $R$ , vertex $C$ is a distance $R$ from both $A$ and $B$ .

Potential (a scalar) adds directly: …
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KEAM 2026Set eng-2026-04214 marksMCQ
$\mu C$
›Reveal solutionSolution

Equal distances $AB=AC$ mean equal potentials at B and C, so no work is done moving a charge between them.

A point charge sits at A of an equilateral triangle, so B and C are at the same distance ( $=$ side $=10$ cm) from it. The potential due to the charge is $V=\dfrac{kQ}{r}$ , identical at B and C:
$V_B=V_C.$
The work to move a charge $q$ from B to C is …
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KEAM 2026Set eng-2026-04214 marksMCQ
$\mu C$
›Reveal solutionSolution

The $V=0$ plane for two equal, opposite charges is the perpendicular bisector, i.e. $0.4$ m from A.

Potential at a point distance $r_1$ from $+q$ and $r_2$ from $-q$ is $V=\dfrac{1}{4\pi\epsilon_0}\left(\dfrac{q}{r_1}-\dfrac{q}{r_2}\right)$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$\mu$
›Reveal solutionSolution

Work done equals the change in electrostatic potential energy of the two point charges as separation goes from 0.4 m to 0.3 m.

Treating each sphere as a point charge: $U=\dfrac{kq^2}{r}$ .

$kq^2=9\times10^9\times(2\times10^{-6})^2=9\times10^9\times4\times10^{-12}=0.036$ . …
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KEAM 2025Set eng-2025-04254 marksMCQ
$A$
›Reveal solutionSolution

Adjusting the charges gives $Q_A = 2C$ and $Q_B = 6C$ ; since $V = Q/C$ with equal $C$ , the potential ratio is $1:3$ .

Each identical capacitor initially carries charge $4C$ and has the same capacitance $C_0$ .

Capacitor A (add $-2C$ ):
$Q_A = 4C - 2C = 2C.$
Capacitor B (add $+2C$ ):

$Q_B = 4C + 2C = 6C.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$Q. The equipotential surface is (A) a plane for a point charge (B) spherical for a dipole (C) cylindrical for a dipole (D) spherical for a point charge (E) cylindrical for a point charge$
›Reveal solutionSolution

A point charge has radial field lines, so its equipotential surfaces are concentric spheres.

The potential of a point charge, $V = \dfrac{kq}{r}$ , depends only on $r$ ; all points at the same $r$ share one potential, forming a spherical surface centred on the charge. These spheres are everywhere perpendicular …
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KEAM 2024Set eng-2024-06084 marksMCQ
$Q. A thin spherical shell of radius 12 cm is charged such that the potential on its surface is 60 V. Then the potential at the centre of the sphere is (A) 5 V (B) Zero (C) 30 V (D) 120 V (E) 60 V$
›Reveal solutionSolution

A charged shell is an equipotential inside; centre potential = surface potential = 60 V.

For a thin charged spherical shell, the electric field inside is zero, so the potential is constant everywhere inside and equal to the surface potential:

$V_{\text{inside}} = V_{\text{surface}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R}.$ …
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KEAM 2022Set eng-2022-P1-A14 marksMCQ
$+5\,\mu$
›Reveal solutionSolution
Moving a charge around a closed loop in an electrostatic field does zero work.

Concept and Intuition

The electrostatic field is conservative, so the work done on a charge over any closed path is zero — the potential returns to its starting value. Carrying $-5\,\mu$ C once around a circle centred on the fixed charge returns it to the same potential.

Step-by-Step Solution

Every point of the circle is at the same distance from the central charge, hence the same potential. …
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Q.Two charges $-q$ each are separated by distance $2d$ . A third charge $+q$ is kept at mid-point $O$ . Find the potential energy of $+q$ as a function of small distance $x$ from $O$ due to $-q$ charges. Sketch P.E. v/s $x$ and convince yourself that the charge at $O$ is in an unstable equilibrium.

Concept understanding — Electric Potential

Electric Potential

The Idea

Potential Due to a Point Charge

Potential Difference

Relation Between Field and Potential

Superposition

Equipotential Surfaces

Method: Testing Equilibrium Stability via Potential-Energy Expansion

Steps

Concept and Intuition

Step-by-Step Solution

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